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Is (10) a solution to x y1?
x+y=1
What is an ordered pair to the following solution -x 2y5 and x y1?
If you mean: -x+2y = 5 and x+y = 1 then their values are x = -1 and y = 2
What is the answer to 3x y1 x y1?
{3x +y =1 {x+y= -3
The distance between the points 2 y1 and 7 8 is 13 What is a possible value of y1?
You can use Pythagoras to solve this to get the co-ordinates. from Pythagoras we have a^2 + b^2 = c^2 For this a and b will be the x and y distance and c the distance between the 2 points so we get (7-2) ^2 + (8-y1)^2 = 13^2 So we need so solve this to get a correct solution which is (8-y1)^2 = 144 so 8 - y1 = + or - 12 so y1 = -4 or y1 = 20
How do you use absolute value to find the distance between two points that have the same x coordinates but different y coordinates?
The distance between (x, y1) and (x, y2) is abs(y1 - y2) or |y1 - y2|.
Determine if the point 3-4 is a solution for the equation y equals 8 -4x?
use the formula y-y1=m(x-x1)
What is a point slope form equation?
(y-y1)=m(x-x1) OR we can write it y=m(x-x1)+y1
What does the point slope formula look like?
y - y1 = m(x - x1), where m is the slope of the line, and (x1, y1) is a point on the line.
What solution is 4x-y1 and -x yx-5?
Without any equality signs the given expressions can't be considered to be equations so therefore there are no solutions.
What is the solution of this question By wronskian determinant method please find whether the solution exist or not where.... y and 8321cos and sup2x y and 83221 plus cos2x?
What is the solution of this question By wronskian determinant method please find whether the solution exist or not where.... y1 = (squre of) cos x, y2 = 1 + cos 2 x.
How do you transfer point slope form into slope intercept form?
First substitute the coordinates of (x1, y1) into the equation, then simplify the equation so it has y in terms of x. y - y1 = m(x - x1) y - y1 = mx - mx1 y = mx - mx1 + y1 y = mx + (y1 - mx1) y = mx + (C)
How do you find linear equations with just coordinates?
if we take the (x1,y1),(x2,y2) as coordinates the formula was (x-x1)/(x2-x1)=(y-y1)/(y2-y1)
